# 68 mL of a 0.28 M CaCl2 solution is added to 92 mL of a 0.46 M CaCl2 solution. Determine the concentration of the combined solution. Show all work with units. How would you solve this?

May 8, 2017

I hope I would solve it correctly.

......I finally get a concentration with respect to $C a C {l}_{2}$ of $\cong 0.4 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

The basic definition of concentration is as amount of solute per unit volume. That is..........

$\text{Concentration"="Moles of solute"/"Volume of solution}$.

Most of the time we want to assess $\text{moles of solute}$, and this is simply the product:

$\text{Moles of solute"="Volume"xx"concentration}$. I would get familiar with this expression, because you will use it a lot.

.........And in problems like these we must assume (REASONABLY!) that the volumes are additive. And so to address your problem (finally!), we solve the quotient:

$\frac{0.092 \cdot L \times 0.46 \cdot m o l \cdot {L}^{-} 1 + 0.068 \cdot L \times 0.28 \cdot m o l \cdot {L}^{-} 1}{\left(92 + 68\right) \times {10}^{-} 3 \cdot L}$

$\cong 0.4 \cdot m o l \cdot {L}^{-} 1$

Do the units in the quotient cancel to give an answer in $m o l \cdot {L}^{-} 1$? It is your problem not mine.

May 8, 2017

The concentration of the combined solution is 0.38 mol/L.

#### Explanation:

1. Calculate the number of moles in each solution.
2. Use the total moles and the total volume to calculate the molarity of the combined solution.

1. Number of moles in each solution

(a) Solution 1

${\text{Moles of CaCl"_2 = 0.068 color(red)(cancel(color(black)("L solution"))) × ("0.28 mol CaCl"_2)/(1 color(red)(cancel(color(black)("L solution")))) = "0.0190 mol CaCl}}_{2}$

(b) Solution2

${\text{Moles of CaCl"_2 = 0.092 color(red)(cancel(color(black)("L solution"))) × ("0.46 mol CaCl"_2)/(1 color(red)(cancel(color(black)("L solution")))) = "0.0423 mol CaCl}}_{2}$

Molarity of combined solutions

$\text{Total moles" = "(0.0190 + 0.0423) mol" = "0.0613 mol}$

$\text{Total volume" = "(68 + 92) mL" = "160 mL" = "0.160 L}$

$\text{Molarity" = "moles"/"litres" = "0.0613 mol"/"0.160 L" = "0.38 mol/L}$